An ideal gas is made to undergo the cyclic pr | vedclass

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(A) In a cyclic process,the system returns to its initial state,so the change in internal energy is zero,i.e.,$\Delta U = 0$.The work done $\Delta W$

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(A) In a cyclic process,the system returns to its initial state,so the change in internal energy is zero,i.e.,$\Delta U = 0$.The work done $\Delta W$ in a $P-V$ diagram is equal to the area enclosed by the cycle. For a clockwise cycle,the work done is positive,and for a counter-clockwise cycle,the work done is negative.Looking at the provided figure,the cycle $A \rightarrow B \rightarrow C \rightarrow A$ is traversed in a counter-clockwise direction. Therefore,the net work done by the gas is negative,i.e.,$\Delta W According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Since $\Delta U = 0$ and $\Delta W Therefore,the signs are $\Delta W < 0$,$\Delta U = 0$,and $Q < 0$. The correct option is $A$.

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